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**1. Homework Statement**

In △ABC, AB=(2-x) cm, BC=(x+1) cm and ∠ABC=120°:

(a) Show that AC^2=x^2-x+7.

(b) Find the value of x for which AC has a minimum value

**3. The Attempt at a Solution**

I am having trouble with part 'b'

am suppose to complete the square , but I dont seem to get the right answer.

[tex] x^2 -x +7 =0 [/tex]

[tex] x^2 -x = -7 [/tex]

[tex] (x- \frac{1}{2} )^2 = -7 + \frac{1}{4} [/tex]

[tex](x- \frac{1}{2} )^2 = \frac{-27}{4} [/tex]

[tex] x- \frac{1}{2} = -\frac{\sqrt27}{2}[/tex]

and than am I suppose to solve for x ? what does it mean by minimum value?